
#1
May2307, 02:39 PM

P: 29

A thin rod 39.2 cm long is charged uniformly with a positive charge density of 72.0 mC/m. The rod is placed along the yaxis and is centered at the origin. A charge of +46.5 mC is placed 68.4 cm from the midpoint of the rod on the positive xaxis. Calculate the electric field at a point on the xaxis, which is halfway between the point charge and the center of the rod. (Express your answer in terms of the unit vector x. For example, if the electric field is 10.5x N/C, then input 10.5 N/C.)
ok..so i drew a picture and i see that the E field of te line charge will be in the x direction becuase that i perpendicular to the line i also see that the point charge will be in the x direction as well becuase it is located on the xaxis so since both charges are postive and the point where i want to know the E field at is between them i am going to subtract their individual E fields at that point to get the net E=ElineEpoint Eline=2K(lamda)/r where r= half the distance betwen the line and point charge Epoint=kq/r^2 so i have E=(2k(lamda)/r)(kq/r^2) does this look right?....if not, please tell me what i am doing wrong 



#2
May2307, 02:57 PM

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#3
May2307, 03:06 PM

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#4
May2307, 03:18 PM

P: 29

electric field of a line charge and point charge
ok so if i am to integrat across the line charge then i would have
integral of Kdq/r^2=kQ/x(xL) where L= the lenght of the rod and x= the distance to the point where i want to find the E field so now i have (KQ/x(xL))(kQpoint/r^2)=E but doesn't gauss's law also work (what i did above) 



#5
May2307, 03:29 PM

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#6
May2307, 03:31 PM

P: 29

i got the simplified answer to the integral from my book....but what about if is had ....
=k/r^2(intergral of dq)=kQ/r^2 does that look better?? 



#7
May2307, 03:39 PM

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#8
May2307, 03:46 PM

P: 29

i see that the equation Ez is the same equation that i wa using in the firs attempt




#9
May2307, 03:46 PM

P: 29

so should Eline=2k(lamda)/r?




#10
May2307, 06:17 PM

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#11
Feb1410, 05:52 PM

P: 138

Ok here can't we just take it as a distance of 68.4/2 = 34.1 cm from the midpoint of the rod? And then just integrate k dq/r^2? with r = 34.1 cm which can be converted to metres later.




#12
Feb1410, 05:59 PM

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#13
Feb1410, 07:47 PM

P: 138

No I am not thinking that
I am thinking that if we take the symmetry from the line charge from both the sides, then we get the x component of the E field at 34.1 cm right? 



#14
Feb1510, 07:21 AM

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#15
Mar110, 08:26 PM

P: 23

you can integrate the function dE=kdQ/(r^2). dQ=lambda dy where lambda is the linear charge density Q/a where a is the length of the line of charge. and you find the y component by multiplying by the unit vector sin(phi) where phi is the angle between the x axis and r where r is the distance from P(x,y) to the upper limit of your line of charge. r=sqrt(x^2+y^2)
SO YOU GET... S dE_y= kQ/(a) S[(y)r^(3/2)]dy. What you will get if you integrate the right side from 0 to a is the y component of your Electric field. To find the x component uses a similar process but the integral is a bit trickier (trig substitution) Then you can plug and chug away. there is probably an easier way to do this but i think the long way helps build an intuition because now its clear as day for me. 


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